Problem: The base of a solid $S$ is the region bounded by the circle $x^2+y^2=16$. $y$ $x$ ${x^2+y^2=16}$ Cross-sections perpendicular to the $x$ -axis are isosceles right triangles with the hypotenuse lying in the base. Determine the exact volume of solid $S$.
Solution: Let's graph the base of the solid. The thin orange rectangle depicts a representative cross-section sitting on the base. The length of the green segment is $2y$. $y$ $x$ $2y$ $(x,y)$ $(x,-y)$ ${x^2+y^2=16}$ Since each cross-section is perpendicular to the $x$ -axis, the independent variable is $x$. If $A$ denotes the area of each cross-section as a function of $x$, the volume $V$ of solid $S$ is $ V=\int_a^b A(x) \,dx$. To determine the area $A$ as a function of $x$, first express $A$ in terms of $y$. Since the triangular cross-section rests on the rectangle pictured above, the length of the base of the triangle is $2y$. Since the triangle is right isosceles with hypotenuse at the base, the height of the triangle is $y$. $y$ $2y$ The area $A$ of the triangle is $A=\dfrac12\cdot2y\cdot y=y^2$. What is $A$ as a function of $x$ ? The corner point $(x,y)$ of the rectangle lies on the circle $x^2+y^2=16$. Let's rewrite the equation as $y^2=16-x^2$. Now we can express $A=y^2$ in terms of $x$ as $A(x)=16-x^2$. Can you express the volume $V$ of solid $S$ as a definite integral? Since $x$ goes from $-4$ to $4$, the volume formula $ V=\int_a^b A(x) \,dx$ gives us the definite integral $ V=\int_{-4}^4 \left(16-x^2\right) dx$. Since we're integrating an even function over a symmetric interval, we can rewrite the integral as $ V=2\int_0^4 \left(16-x^2\right) dx$. What is the value of the integral? $\begin{aligned} V&=2\int_0^4 \left(16-x^2\right) dx \\\\ &=2\left[16x-\dfrac13x^3\right]_0^4 \\\\ &=2\left[16(4)-\dfrac13(4)^3-\left(16(0)-\dfrac13(0)^3\right)\right] \\\\ &=2\left[64-\dfrac{64}3\right] \\\\ &=\dfrac{256}3 \end{aligned}$